📚 Mastering Calculation Questions from A-Level Chemistry June 2018 Insert 3 | 掌握 2018 年 6 月 A-Level 化学插入材料 3 的计算题型
The June 2018 A-Level Chemistry Insert 3 presents a range of experimental data around a key reaction, typically challenging students with integrated calculation tasks. This article dissects the step‑by‑step approaches required to tackle rate equations, activation energy, equilibrium constants, titration‑based moles, and percentage yield. By working through realistic worked examples linked to the insert, you will gain confidence in handling similar multi‑step problems under exam conditions.
2018 年 6 月的 A‑Level 化学插入材料 3 提供了一系列围绕某一核心反应的实验数据，通常要求考生综合运用多种计算技能。本文逐步拆解处理速率方程、活化能、平衡常数、滴定求摩尔以及产率计算所需的解题思路。通过完成与插入材料相关的真实范例，你将更自信地在考试中应对同类多步计算题。
1. Deciphering the Insert Data | 解读插入材料中的数据
The insert for June 2018 typically includes tables of initial rates at varying concentrations, measurements of gas volumes, temperature‑dependent rate constants, and titration results. The first step is to identify exactly which quantities are being measured and their units. For instance, an initial‑rate table might list [NO] in mol dm⁻³, [H₂] in mol dm⁻³, and initial rate in mol dm⁻³ s⁻¹. Always circle the headings and note whether the data come from the same experiment or different runs.
2018 年 6 月的插入材料通常包含不同浓度下的初始速率表格、气体体积测量值、随温度变化的速率常数以及滴定结果。第一步是准确识别被测量的量及其单位。例如，初始速率表格可能列出 NO 的浓度（mol dm⁻³）、H₂ 的浓度（mol dm⁻³）和初始速率（mol dm⁻³ s⁻¹）。始终圈出表头，并注意数据是来自同一实验还是不同实验批次。
| Experiment | [NO] / mol dm⁻³ | [H₂] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.5 × 10⁻³ |
| 2 | 0.20 | 0.10 | 5.0 × 10⁻³ |
| 3 | 0.20 | 0.20 | 2.0 × 10⁻² |
2. Determining Reaction Orders from Initial Rates | 通过初始速率确定反应级数
reaction is first order with respect to NO. Now compare Experiments 2 and 3: [NO] stays at 0.20 mol dm⁻³, while [H₂] doubles from 0.10 to 0.20 mol dm⁻³. The rate increases four‑fold (5.0 × 10⁻³ → 2.0 × 10⁻²), so the reaction is second order with respect to H₂.
比较仅一个反应物浓度发生变化的实验对。实验 1 与 2 中，[H₂] 恒定，[NO] 加倍（0.10 → 0.20 mol dm⁻³），速率也加倍（2.5 × 10⁻³ → 5.0 × 10⁻³），表明对 NO 为一级。再比较实验 2 和 3：[NO] 保持 0.20 mol dm⁻³，[H₂] 从 0.10 加倍到 0.20 mol dm⁻³，速率增大四倍（5.0 × 10⁻³ → 2.0 × 10⁻²），因此对 H₂ 为二级。整体反应为三级。
Rate ∝ [NO]¹[H₂]² → Rate = k[NO][H₂]²
3. Writing and Justifying the Rate Equation | 书写并验证速率方程
¹ = 2, so order=1. For Experiment 3 ÷ Experiment 2, rate ratio = (2.0×10⁻²)/(5.0×10⁻³)=4, concentration ratio for H₂ = 0.20/0.10=2 → 2² = 4, so order=2.
利用推导出的级数，速率方程为 rate = k[NO][H₂]²。必须用配对数据证明每个级数，而非仅仅陈述结论。用数学形式展示比值：实验 2 除以实验 1，速率比 = (5.0×10⁻³)/(2.5×10⁻³)=2，NO 浓度比 = 0.20/0.10=2 → 2¹ = 2，故级数=1。实验 3 除以实验 2，速率比 = (2.0×10⁻²)/(5.0×10⁻³)=4，H₂ 浓度比 = 0.20/0.10=2 → 2² = 4，故级数=2。整体为三级反应。
4. Calculating the Rate Constant k | 计算速率常数 k
ince overall order=3, k = rate / (concentration)³ → units = (mol dm⁻³ s⁻¹) / (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
确定速率方程后，代入任意一组实验数据即可求出 k。使用实验 1：速率 = 2.5 × 10⁻³ mol dm⁻³ s⁻¹，[NO]=0.10，[H₂]=0.10。k = 速率 / ([NO][H₂]²) = (2.5 × 10⁻³) / (0.10 × (0.10)²) = (2.5 × 10⁻³) / 1.0 × 10⁻³ = 2.5。由于总级数为三级，k 的单位 = (mol dm⁻³ s⁻¹) / (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹。记得注明单位。
k = 2.5 dm⁶ mol⁻² s⁻¹
5. Using the Arrhenius Equation | 运用阿伦尼乌斯方程
k against 1/T yields a straight line with slope = −Eₐ/R. Alternatively, if two temperature‑rate constant pairs are given, use the two‑point form: ln(k₂/k₁) = −Eₐ/R (1/T₂ − 1/T₁). R = 8.31 J K⁻¹ mol⁻¹.
插入材料常给出不同温度下的速率常数，用来计算活化能 Eₐ。阿伦尼乌斯方程的对数形式为 ln k = ln A − Eₐ/(RT)。以 ln k 对 1/T 作图可得直线，斜率 = −Eₐ/R。如果给出两个温度下的 k，可用两点式：ln(k₂/k₁) = −Eₐ/R (1/T₂ − 1/T₁)。通用气体常数 R = 8.31 J K⁻¹ mol⁻¹。注意温度单位需转换为开尔文。
Example: n(k₂/k₁)) / (1/T₂ − 1/T₁) = −8.31 × 1.386 / (−1.30×10⁻⁴) ≈ 88 500 J mol⁻¹ = 88.5 kJ mol⁻¹.
例子：298 K 时 k₁ = 2.5 × 10⁻³；310 K 时 k₂ = 1.0 × 10⁻²。ln(k₂/k₁) = ln(4) = 1.386。1/T₂ − 1/T₁ = (1/310 − 1/298) = −1.30 × 10⁻⁴ K⁻¹。Eₐ = −8.31 × 1.386 / (−1.30×10⁻⁴) ≈ 88 500 J mol⁻¹ = 88.5 kJ mol⁻¹。务必展示完整的代入和单位处理。
6. Solving Equilibrium Constant Problems | 解答平衡常数问题
ach involves an ICE table (Initial, Change, Equilibrium). For the reaction 2NO(g) + 2H₂(g) ⇌ 0.10)(0.20)² / (0.30²×0.30²) = 0.004 / (0.00243) = 1.65 mol⁻¹ dm³ (check units carefully).
插入材料可能提供气相或溶液平衡的初始和平衡物质的量。经典方法使用 ICE 表格（初始、变化、平衡）。对于反应 2NO(g) + 2H₂(g) ⇌ N₂(g) + 2H₂O(g)，假设在 1.0 dm³ 容器中起始时 NO 和 H₂ 均为 0.50 mol。平衡时存在 0.10 mol N₂。N₂ 变化为 +0.10，则 NO 减少 0.20，H₂ 减少 0.20；平衡量：NO = 0.30，H₂ = 0.30，H₂O = 0.20。Kc = [N₂][H₂O]²/([NO]²[H₂]²) = 0.10×(0.20)²/(0.30²×0.30²) = 0.004/0.00243 ≈ 1.65（单位：mol⁻¹ dm³）。务必检查 Kc 的单位是否随反应式变化。
7. Back‑Titration for Purity Determination | 返滴定法测定纯度
Titration data in the insert often relate to a back‑ 0500 mol. Moles of NaOH used = 0.0250 dm³ × 0.500 = 0.0125 mol, which neutralises the same moles of HCl leftover. Moles of HCl that reacted = 0.0500 − 0.0125 = 0.0375 mol. The reaction CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O shows a 1:2 mole ratio, so moles of CaCO₃ = 0.0375/2 = 0.01875 mol. Mass = 0.01875 × 100.1 = 1.88 g. If the sample weighed 2.00 g, purity = (1.88/2.00)×100 = 94.0%.
插入材料中的滴定数据常设定为返滴定：加入过量试剂，再滴定剩余部分。举例：某不纯 CaCO₃ 样品与 50.0 cm³ 1.00 mol dm⁻³ HCl（过量）反应，剩余的 HCl 需 25.0 cm³ 0.500 mol dm⁻³ NaOH 中和。原 HCl 物质的量 = 0.0500 × 1.00 = 0.0500 mol。NaOH 物质的量 = 0.0250 × 0.500 = 0.0125 mol，等同于剩余 HCl 的物质的量。参与反应的 HCl = 0.0500 − 0.0125 = 0.0375 mol。CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O，摩尔比 1:2，CaCO₃ 物质的量 = 0.0375/2 = 0.01875 mol，质量 = 0.01875 × 100.1 = 1.88 g。若样品称重 2.00 g，纯度 = (1.88/2.00)×100 = 94.0%。
8. Calculating Percentage Yield in a Multi‑Step Synthesis | 多步合成中的产率计算
A percentage‑ (1 mol) is 80.0 g NH₄NO₃, but only 70.0 g collected. Step 1 yield = 70.0/80.0 × 100 = 87.5% .0 g from that 70.0 g, but only 35.0 g N₂O obtained. Step 2 yield = 35.0/44.0 × 100 = 79.5%. Overall yield = (87.5/100) × (79.5/100) × 100 = 69.6% eoretical mass from 17.0 g NH₃ would be 44.0 g N₂O (via NH₄NO₃), so 35.0/44.0 × 100 = 79.5% if assuming Step 1 is quantitative? and then to N₂O), but Step 1 had a loss. So the overall yield = (35.0 / 44.0) × 100 = 79.5%? if I only obtained 35.0 g N₂O from 17.0 g NH₃, the overall yield = 35.0/44.0 × 100 = 79.5%. But wait, 70.0 g NH₄NO₃ corresponds to 0.875 mol, and from 0.875 mol NH₄NO₃ you' g theoretical in Step 2. But they obtained 35.0 g. So overall yield = (35.0 / 44.0) = 79.5% he theoretical from NH₃ is 44.0 g. However, the yield in Step 2 should be 35.0/38.5 = 90.9%. There' limiting, 17.0 g (1 mol) → theoretical N₂O = 44.0 g. Actual = 35.0 g. Overall yield = 79.5% the consistent approach: overall yield = (actual final / theoretical from start) × 100. We'll clarify.
产率计算题可能涉及两步反应的理质量与实际量。假设第一步：NH₃ + HNO₃ → NH₄NO₃，从 17.0 g NH₃（1 mol）得到理质量 80.0 g NH₄NO₃，但实际只收集到 70.0 g。第一步产率 = 70.0/80.0 × 100 = 87.5%。第二步用 NH₄NO₃ 制 N₂O，从 70.0 g 理质量可得 44.0 g N₂O，但实际仅获得 35.0 g。第二步产率 = 35.0/44.0 × 100 = 79.5%。总产率 = (87.5/100)×(79.5/100)×100 = 69.6%。另一种方法：若基于起始 NH₃ 计算理质量 N₂O 为 44.0 g，则总产率 = 35.0/44.0 × 100 = 79.5%（此处有出入）。正确做法必须明确理论产量基准：多步合成中总产率 =（终产物实际质量 / 由初始原料直接计算的理论质量）× 100%。因此应为 35.0/44.0 × 100 = 79.5%，而非简单乘积分步产率。务必根据题目要求选择合适方法，通常考试推荐使用初始原料理论质量来算总产率。
9. Gas Volume Calculations with Molar Volume | 结合气体摩尔体积的计算
. Volume of O₂ at RTP = 0.00500 × 24.0 = 0.120 dm³ = 120 cm³. Care must be taken with unit conversions: 1 dm³ = 1000 cm³.
插入材料可能包含在室温常压（RTP，摩尔体积≈24.0 dm³ mol⁻¹）或标准状况（STP，22.4 dm³ mol⁻¹）下的气体体积数据。对于反应 2H₂O₂ → 2H₂O + O₂，若 50.0 cm³ 0.200 mol dm⁻³ H₂O₂ 完全分解，H₂O₂ 物质的量 = 0.0500 × 0.200 = 0.0100 mol。O₂ 物质的量 = 0.0100/2 = 0.00500 mol。在 RTP 下 O₂ 体积 = 0.00500 × 24.0 = 0.120 dm³ = 120 cm³。务必注意单位换算：1 dm³ = 1000 cm³，避免数量级错误。
10. Combining Mole Concepts with Enthalpy Calculations | 焓变与摩尔综合计算
= 4.18 J g⁻¹ K⁻¹. Heat released q = mcΔT = 100 × 4.18 × 6.5 = 2717 J. Moles of HCl = 0.0500 dm³ × 1.00 = 0.0500 mol. Enthalpy change per mole ΔH = −q/n = −2717 J / 0.0500 mol = −54 340 J mol⁻¹ = −54.3 kJ mol⁻¹. Pay attention to the sign (exothermic → negative).
插入材料的量热数据可能需要计算每摩尔的焓变。假设 50.0 cm³ 1.00 mol dm⁻³ HCl 与 50.0 cm³ 1.00 mol dm⁻³ NaOH 混合，温度升高 ΔT = 6.5 °C。总体积 = 100 cm³，溶液质量 ≈ 100 g（设密度 1 g cm⁻³），比热容 c = 4.18 J g⁻¹ K⁻¹。放热量 q = mcΔT = 100 × 4.18 × 6.5 = 2717 J。HCl 物质的量 = 0.0500 × 1.00 = 0.0500 mol。每摩尔焓变 ΔH = −q/n = −2717 J / 0.0500 mol = −54 340 J mol⁻¹ = −54.3 kJ mol⁻¹。注意放热反应加负号。
11. Summary of Essential Formulae | 必备公式速览
- Rate equation: rate = k[A]ᵐ[B]ⁿ
- Arrhenius: ln k = ln A − Eₐ/(RT) or ln(k₂/k₁) = −Eₐ/R (1/T₂ − 1/T₁)
- Kc: Kc = [products]ᵖ / [reactants]ʳ
- Moles from solution: n = cV (V in dm³)
- Moles from gas at RTP: n = V (dm³) / 24.0
- Heat change: q = mcΔT; ΔH = −q/n
- Percentage yield: % = (actual mass / theoretical mass) × 100
- Percentage purity: % = (mass of pure / mass of sample) × 100
同样的公式中文版：速率方程、阿伦尼乌斯方程、平衡常数 Kc、溶液物质的量 n=cV、RTP 下气体物质的量 n=V/24.0、热量变化 q=mcΔT 及 ΔH=−q/n、产率 = (实际质量/理论质量)×100%、纯度 = (纯物质质量/样品质量)×100%。牢记这些公式并注意单位换算，是应对插入材料计算的基础。
12. Common Pitfalls and How to Avoid Them | 常见失分点与对策
Examiners' e overall order matches the units of k, and verify that temperature is in Kelvin. For ICE tables, ensure changes correspond correctly to the stoichiometry. Finally, double‑ck that enthalpy signs are consistent with temperature changes (exothermic = negative ΔH).
阅卷报告反复强调的典型错误包括：计算 k 时忘记对浓度取平方或立方，在阿伦尼乌斯计算中用摄氏温度而非开尔文，摩尔比中小数点错位，混淆 dm³ 与 cm³。必须在每一步计算中都标明单位，检查总级数与 k 的单位是否匹配，确认温度已转换为开尔文。使用 ICE 表格时，确保变化量符合计量比。最后，复查焓变符号是否与温度升降一致（放热对应负值 ΔH）。
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